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vampirchik [111]

2 years ago

9

The following three shapes are based only on squares, semicircles, and quarter circles. Find the perimeter and the area of each

shaded part. Give your answer as a completely simplified exact value in terms of π (no approximations).

1 answer:

Colt1911 [192]2 years ago

7 0

The answer is 2.5 because you need to divide it and then you get your answer

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A square serving platter has an area of 124 inches squared. Which is the best estimate of the side length of the platter?

mafiozo [28]

Answer:

Approximately 11 inches

Step-by-step explanation:

7 0

3 years ago

1. Express <img src="https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7Bx%282x%2B3%29%20%7D" id="TexFormula1" title="\frac{1}{x(2x+3) }" a

katovenus [111]

1. Let a and b be coefficients such that

$\dfrac1{x(2x+3)} = \dfrac ax + \dfrac b{2x+3}$

Combining the fractions on the right gives

$\dfrac1{x(2x+3)} = \dfrac{a(2x+3) + bx}{x(2x+3)}$

$\implies 1 = (2a+b)x + 3a$

$\implies \begin{cases}3a=1 \\ 2a+b=0\end{cases} \implies a=\dfrac13, b = -\dfrac23$

so that

$\dfrac1{x(2x+3)} = \boxed{\dfrac13 \left(\dfrac1x - \dfrac2{2x+3}\right)}$

2. a. The given ODE is separable as

$x(2x+3) \dfrac{dy}dx} = y \implies \dfrac{dy}y = \dfrac{dx}{x(2x+3)}$

Using the result of part (1), integrating both sides gives

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + C$

Given that y = 1 when x = 1, we find

$\ln|1| = \dfrac13 \left(\ln|1| - \ln|5|\right) + C \implies C = \dfrac13\ln(5)$

so the particular solution to the ODE is

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3|\right) + \dfrac13\ln(5)$

We can solve this explicitly for y :

$\ln|y| = \dfrac13 \left(\ln|x| - \ln|2x+3| + \ln(5)\right)$

$\ln|y| = \dfrac13 \ln\left|\dfrac{5x}{2x+3}\right|$

$\ln|y| = \ln\left|\sqrt[3]{\dfrac{5x}{2x+3}}\right|$

$\boxed{y = \sqrt[3]{\dfrac{5x}{2x+3}}}$

2. b. When x = 9, we get

$y = \sqrt[3]{\dfrac{45}{21}} = \sqrt[3]{\dfrac{15}7} \approx \boxed{1.29}$

8 0

2 years ago

An electrical engineer wishes to compare the mean lifetimes of two types of transistors in an application involving high-tempera

Verizon [17]

Answer:

add the decimals thats all

Step-by-step explanation:

7 0

3 years ago

Trains headed for destination A arrive at the train station at 15-minute intervals tarting at 7am, where as trains headed for de

grandymaker [24]

Answer: a.) 2/3 ; b.) 7/12

Step-by-step explanation:

Given the following details :

Destination A trains arrives at station at 15 minutes interval starting from 7am

Destination B trains arrives at station at 15 minutes interval starting from 7.05am

A.) if a passenger arrives at station uniformly between 7am and 8am and gets on first train. What proportion of time does the passenger leave from destination A?

To leave for destination A. The passenger has to arrive before train to A departs and after train to B leaves.

Train to B leaves, starting at at 7.05 a.m and at interval of 15 minutes

Train to A leaves at 7.00 am and at interval of 15 minutes.

Therefore,

(Train B departs - train A arrives) (between 7 to 8am)

(7.05 - 7.15) + (7.20 - 7.30) + (7.35 - 7.45) + (7.50 - 8.00) = 40 minutes

Total time = 7.00 - 8.00 = 60 minutes

(40 ÷ 60) = 2/3 or 0.667

B.) If passenger arrives uniformly between 7.10am - 8:10am

(7.10 - 7.15) + (7.20 - 7.30) + (7.35 - 7.45) + (7.50 - 8)

5 + 10 + 10 + 10 = 35

35/60 = 7/12 or 0.583

8 0

3 years ago

What is this in simplest form?

katrin [286]

40/50

80 divided by 2
100 divided by 2

5 0

2 years ago